QBManager

Application of Derivatives

85401 If $x + y = k$ is normal to $y^{2} = 12 x$ , then $k$ is equal to

1 3

2 9

3 -9

4 -3

Explanation:

(B) : Given,
$y^{2} = 12 x$
On differentiating w.r.t $x$ , we get -
$2 y \frac{d y}{d x} = 12 \Rightarrow \frac{d y}{d x} = \frac{12}{2 y}$
$\frac{d y}{d x} = \frac{6}{y}$
$∴$ Slope of the normal $= - \frac{1}{\frac{dy}{dx}} = \frac{- y}{6}$
Slope of the line $x + y = k$ is -1
$\frac{- y}{6} = - 1$
$y = 6$
Putting the value of $y = 6$ in equation (i), we get $6^{2} = 12 x$
$36 = 12 x$
$x = \frac{36}{12}$
$x = 3$
Putting the value of $x$ and $y$ in equation $x + y = k$
$x + y = k$
$3 + 6 = k$
$k = 9$

MHT CET-2011

Application of Derivatives

85402 Length of the sub tangent at $(a, a)$ on the curve
$y^{2} = \frac{x^{2}}{2 a + x} is equal to$

1

\frac{18}{5}

2

\frac{18 a}{5}

3

\frac{18 a^{2}}{5}

4

- \frac{18 a^{2}}{5}

Explanation:

(C) : Given,
On differentiating both sides,
$2 y \frac{d y}{d x} = \frac{(2 a + x) (2 x) - x^{2} (1)}{(2 a + x)^{2}}$
${[\frac{d y}{d x}]}_{(a, a)} = \frac{(2 a + a) (2 a) - a^{2}}{(2 a + a)^{2}}$
${[\frac{d y}{d x}]}_{(a, a)} = \frac{(3 a) (2 a) - a^{2}}{2 a (3 a)^{2}} = \frac{6 a^{2} - a^{2}}{2 a \times 9 a^{2}} = \frac{5 a^{2}}{18 a^{3}}$
Length of subtangent at $(a, a)$ ,
$= \frac{y}{{(\frac{d y}{d x})}_{(a, a)}} = \frac{a}{\frac{5}{18 a}} = a \times \frac{18 a}{5} = \frac{18 a^{2}}{5}$

COMEDK-2012

Application of Derivatives

85403 If the tangent to the curve $2 y^{3} = a x^{2} + x^{3}$ at the point ( $a$ , a) cuts off intercepts $α$ and $β$ on the coordinate axes where $α^{2} + β^{2} = 61$ , then the value of $a$ is

1 25

2 36

3

\pm 30

4

\pm 40

Explanation:

(C) : Given, the curve -
$2 y^{3} = a x^{2} + x^{3}$
On differentiating both sides w.r.t. $x$ we get -
$6 y^{2} \frac{d y}{d x} = 2 a x + 3 x^{2}$
$\frac{d y}{d x} = \frac{2 a x + 3 x^{2}}{6 y^{2}}$
${(\frac{d y}{d x})}_{(a, a)} = \frac{5 a^{2}}{6 a^{2}}$
${(\frac{d y}{d x})}_{(a, a)} = \frac{5}{6}$
The equation of the tangent at $(a, a)$ is,
$y - a = \frac{5}{6} (x - a)$
$6 y - 6 a = 5 x - 5 a$
$5 x - 6 y + a = 0$
This intercepts lengths $= \frac{- a}{5}$ and $\frac{a}{6}$ with $x$ and $y$ axis respectively.
$∴ α = \frac{- a}{5}, β = \frac{a}{6}$
Now, $α^{2} + β^{2} = 61$
${(\frac{- a}{5})}^{2} + {(\frac{a}{6})}^{2} = 61$
$\frac{a^{2}}{25} + \frac{a^{2}}{36} = 61$
$\frac{36 a^{2} + 25 a^{2}}{25 \times 36} = 61$
$61 a^{2} = 61 \times 36 \times 25$
$a^{2} = 36 \times 25$
$a = 6 \times 5$
$a = \pm 30$
The value of a is $\pm 30$ .

COMEDK-2012

Application of Derivatives

85404 If the radius of a sphere is measured as $9 cm$ with an error of $0.03 cm$ , then find the approximating error in calculating its volume.

1

2.46 π {cm}^{3}

2

8.62 π {cm}^{3}

3

9.72 π {cm}^{3}

4

7.46 π {cm}^{3}

Explanation:

(C) : We have,
Radius of sphere $(r) = 9 cm$
and Error $Δ r = 0.03 cm$
Volume of sphere,
$V = \frac{4}{3} π r^{3}$
$\frac{dV}{dr} = 4 π r^{2}$
${(\frac{dV}{dr})}_{(r = 9)} = 4 π \times 9^{2} = 324 π$
Let, $Δ V$ be the error in $V$ due to error $Δ r$ in $r$ .
Then, $Δ V = \frac{dV}{dr} Δ r$
$Δ V = 324 π \times 0.03 = 9.72 π {cm}^{3}$
Thus, the approximate error in calculating the volume is $9.72 π {cm}^{3}$ .

COMEDK-2016

Application of Derivatives

85401 If $x + y = k$ is normal to $y^{2} = 12 x$ , then $k$ is equal to

1 3

2 9

3 -9

4 -3

Explanation:

(B) : Given,
$y^{2} = 12 x$
On differentiating w.r.t $x$ , we get -
$2 y \frac{d y}{d x} = 12 \Rightarrow \frac{d y}{d x} = \frac{12}{2 y}$
$\frac{d y}{d x} = \frac{6}{y}$
$∴$ Slope of the normal $= - \frac{1}{\frac{dy}{dx}} = \frac{- y}{6}$
Slope of the line $x + y = k$ is -1
$\frac{- y}{6} = - 1$
$y = 6$
Putting the value of $y = 6$ in equation (i), we get $6^{2} = 12 x$
$36 = 12 x$
$x = \frac{36}{12}$
$x = 3$
Putting the value of $x$ and $y$ in equation $x + y = k$
$x + y = k$
$3 + 6 = k$
$k = 9$

MHT CET-2011

Application of Derivatives

85402 Length of the sub tangent at $(a, a)$ on the curve
$y^{2} = \frac{x^{2}}{2 a + x} is equal to$

1

\frac{18}{5}

2

\frac{18 a}{5}

3

\frac{18 a^{2}}{5}

4

- \frac{18 a^{2}}{5}

Explanation:

(C) : Given,
On differentiating both sides,
$2 y \frac{d y}{d x} = \frac{(2 a + x) (2 x) - x^{2} (1)}{(2 a + x)^{2}}$
${[\frac{d y}{d x}]}_{(a, a)} = \frac{(2 a + a) (2 a) - a^{2}}{(2 a + a)^{2}}$
${[\frac{d y}{d x}]}_{(a, a)} = \frac{(3 a) (2 a) - a^{2}}{2 a (3 a)^{2}} = \frac{6 a^{2} - a^{2}}{2 a \times 9 a^{2}} = \frac{5 a^{2}}{18 a^{3}}$
Length of subtangent at $(a, a)$ ,
$= \frac{y}{{(\frac{d y}{d x})}_{(a, a)}} = \frac{a}{\frac{5}{18 a}} = a \times \frac{18 a}{5} = \frac{18 a^{2}}{5}$

COMEDK-2012

Application of Derivatives

85403 If the tangent to the curve $2 y^{3} = a x^{2} + x^{3}$ at the point ( $a$ , a) cuts off intercepts $α$ and $β$ on the coordinate axes where $α^{2} + β^{2} = 61$ , then the value of $a$ is

1 25

2 36

3

\pm 30

4

\pm 40

Explanation:

(C) : Given, the curve -
$2 y^{3} = a x^{2} + x^{3}$
On differentiating both sides w.r.t. $x$ we get -
$6 y^{2} \frac{d y}{d x} = 2 a x + 3 x^{2}$
$\frac{d y}{d x} = \frac{2 a x + 3 x^{2}}{6 y^{2}}$
${(\frac{d y}{d x})}_{(a, a)} = \frac{5 a^{2}}{6 a^{2}}$
${(\frac{d y}{d x})}_{(a, a)} = \frac{5}{6}$
The equation of the tangent at $(a, a)$ is,
$y - a = \frac{5}{6} (x - a)$
$6 y - 6 a = 5 x - 5 a$
$5 x - 6 y + a = 0$
This intercepts lengths $= \frac{- a}{5}$ and $\frac{a}{6}$ with $x$ and $y$ axis respectively.
$∴ α = \frac{- a}{5}, β = \frac{a}{6}$
Now, $α^{2} + β^{2} = 61$
${(\frac{- a}{5})}^{2} + {(\frac{a}{6})}^{2} = 61$
$\frac{a^{2}}{25} + \frac{a^{2}}{36} = 61$
$\frac{36 a^{2} + 25 a^{2}}{25 \times 36} = 61$
$61 a^{2} = 61 \times 36 \times 25$
$a^{2} = 36 \times 25$
$a = 6 \times 5$
$a = \pm 30$
The value of a is $\pm 30$ .

COMEDK-2012

Application of Derivatives

85404 If the radius of a sphere is measured as $9 cm$ with an error of $0.03 cm$ , then find the approximating error in calculating its volume.

1

2.46 π {cm}^{3}

2

8.62 π {cm}^{3}

3

9.72 π {cm}^{3}

4

7.46 π {cm}^{3}

Explanation:

(C) : We have,
Radius of sphere $(r) = 9 cm$
and Error $Δ r = 0.03 cm$
Volume of sphere,
$V = \frac{4}{3} π r^{3}$
$\frac{dV}{dr} = 4 π r^{2}$
${(\frac{dV}{dr})}_{(r = 9)} = 4 π \times 9^{2} = 324 π$
Let, $Δ V$ be the error in $V$ due to error $Δ r$ in $r$ .
Then, $Δ V = \frac{dV}{dr} Δ r$
$Δ V = 324 π \times 0.03 = 9.72 π {cm}^{3}$
Thus, the approximate error in calculating the volume is $9.72 π {cm}^{3}$ .

COMEDK-2016

Application of Derivatives

85401 If $x + y = k$ is normal to $y^{2} = 12 x$ , then $k$ is equal to

1 3

2 9

3 -9

4 -3

Explanation:

(B) : Given,
$y^{2} = 12 x$
On differentiating w.r.t $x$ , we get -
$2 y \frac{d y}{d x} = 12 \Rightarrow \frac{d y}{d x} = \frac{12}{2 y}$
$\frac{d y}{d x} = \frac{6}{y}$
$∴$ Slope of the normal $= - \frac{1}{\frac{dy}{dx}} = \frac{- y}{6}$
Slope of the line $x + y = k$ is -1
$\frac{- y}{6} = - 1$
$y = 6$
Putting the value of $y = 6$ in equation (i), we get $6^{2} = 12 x$
$36 = 12 x$
$x = \frac{36}{12}$
$x = 3$
Putting the value of $x$ and $y$ in equation $x + y = k$
$x + y = k$
$3 + 6 = k$
$k = 9$

MHT CET-2011

Application of Derivatives

85402 Length of the sub tangent at $(a, a)$ on the curve
$y^{2} = \frac{x^{2}}{2 a + x} is equal to$

1

\frac{18}{5}

2

\frac{18 a}{5}

3

\frac{18 a^{2}}{5}

4

- \frac{18 a^{2}}{5}

Explanation:

(C) : Given,
On differentiating both sides,
$2 y \frac{d y}{d x} = \frac{(2 a + x) (2 x) - x^{2} (1)}{(2 a + x)^{2}}$
${[\frac{d y}{d x}]}_{(a, a)} = \frac{(2 a + a) (2 a) - a^{2}}{(2 a + a)^{2}}$
${[\frac{d y}{d x}]}_{(a, a)} = \frac{(3 a) (2 a) - a^{2}}{2 a (3 a)^{2}} = \frac{6 a^{2} - a^{2}}{2 a \times 9 a^{2}} = \frac{5 a^{2}}{18 a^{3}}$
Length of subtangent at $(a, a)$ ,
$= \frac{y}{{(\frac{d y}{d x})}_{(a, a)}} = \frac{a}{\frac{5}{18 a}} = a \times \frac{18 a}{5} = \frac{18 a^{2}}{5}$

COMEDK-2012

Application of Derivatives

85403 If the tangent to the curve $2 y^{3} = a x^{2} + x^{3}$ at the point ( $a$ , a) cuts off intercepts $α$ and $β$ on the coordinate axes where $α^{2} + β^{2} = 61$ , then the value of $a$ is

1 25

2 36

3

\pm 30

4

\pm 40

Explanation:

(C) : Given, the curve -
$2 y^{3} = a x^{2} + x^{3}$
On differentiating both sides w.r.t. $x$ we get -
$6 y^{2} \frac{d y}{d x} = 2 a x + 3 x^{2}$
$\frac{d y}{d x} = \frac{2 a x + 3 x^{2}}{6 y^{2}}$
${(\frac{d y}{d x})}_{(a, a)} = \frac{5 a^{2}}{6 a^{2}}$
${(\frac{d y}{d x})}_{(a, a)} = \frac{5}{6}$
The equation of the tangent at $(a, a)$ is,
$y - a = \frac{5}{6} (x - a)$
$6 y - 6 a = 5 x - 5 a$
$5 x - 6 y + a = 0$
This intercepts lengths $= \frac{- a}{5}$ and $\frac{a}{6}$ with $x$ and $y$ axis respectively.
$∴ α = \frac{- a}{5}, β = \frac{a}{6}$
Now, $α^{2} + β^{2} = 61$
${(\frac{- a}{5})}^{2} + {(\frac{a}{6})}^{2} = 61$
$\frac{a^{2}}{25} + \frac{a^{2}}{36} = 61$
$\frac{36 a^{2} + 25 a^{2}}{25 \times 36} = 61$
$61 a^{2} = 61 \times 36 \times 25$
$a^{2} = 36 \times 25$
$a = 6 \times 5$
$a = \pm 30$
The value of a is $\pm 30$ .

COMEDK-2012

Application of Derivatives

85404 If the radius of a sphere is measured as $9 cm$ with an error of $0.03 cm$ , then find the approximating error in calculating its volume.

1

2.46 π {cm}^{3}

2

8.62 π {cm}^{3}

3

9.72 π {cm}^{3}

4

7.46 π {cm}^{3}

Explanation:

(C) : We have,
Radius of sphere $(r) = 9 cm$
and Error $Δ r = 0.03 cm$
Volume of sphere,
$V = \frac{4}{3} π r^{3}$
$\frac{dV}{dr} = 4 π r^{2}$
${(\frac{dV}{dr})}_{(r = 9)} = 4 π \times 9^{2} = 324 π$
Let, $Δ V$ be the error in $V$ due to error $Δ r$ in $r$ .
Then, $Δ V = \frac{dV}{dr} Δ r$
$Δ V = 324 π \times 0.03 = 9.72 π {cm}^{3}$
Thus, the approximate error in calculating the volume is $9.72 π {cm}^{3}$ .

COMEDK-2016

Application of Derivatives

85401 If $x + y = k$ is normal to $y^{2} = 12 x$ , then $k$ is equal to

1 3

2 9

3 -9

4 -3

Explanation:

(B) : Given,
$y^{2} = 12 x$
On differentiating w.r.t $x$ , we get -
$2 y \frac{d y}{d x} = 12 \Rightarrow \frac{d y}{d x} = \frac{12}{2 y}$
$\frac{d y}{d x} = \frac{6}{y}$
$∴$ Slope of the normal $= - \frac{1}{\frac{dy}{dx}} = \frac{- y}{6}$
Slope of the line $x + y = k$ is -1
$\frac{- y}{6} = - 1$
$y = 6$
Putting the value of $y = 6$ in equation (i), we get $6^{2} = 12 x$
$36 = 12 x$
$x = \frac{36}{12}$
$x = 3$
Putting the value of $x$ and $y$ in equation $x + y = k$
$x + y = k$
$3 + 6 = k$
$k = 9$

MHT CET-2011

Application of Derivatives

85402 Length of the sub tangent at $(a, a)$ on the curve
$y^{2} = \frac{x^{2}}{2 a + x} is equal to$

1

\frac{18}{5}

2

\frac{18 a}{5}

3

\frac{18 a^{2}}{5}

4

- \frac{18 a^{2}}{5}

Explanation:

(C) : Given,
On differentiating both sides,
$2 y \frac{d y}{d x} = \frac{(2 a + x) (2 x) - x^{2} (1)}{(2 a + x)^{2}}$
${[\frac{d y}{d x}]}_{(a, a)} = \frac{(2 a + a) (2 a) - a^{2}}{(2 a + a)^{2}}$
${[\frac{d y}{d x}]}_{(a, a)} = \frac{(3 a) (2 a) - a^{2}}{2 a (3 a)^{2}} = \frac{6 a^{2} - a^{2}}{2 a \times 9 a^{2}} = \frac{5 a^{2}}{18 a^{3}}$
Length of subtangent at $(a, a)$ ,
$= \frac{y}{{(\frac{d y}{d x})}_{(a, a)}} = \frac{a}{\frac{5}{18 a}} = a \times \frac{18 a}{5} = \frac{18 a^{2}}{5}$

COMEDK-2012

Application of Derivatives

85403 If the tangent to the curve $2 y^{3} = a x^{2} + x^{3}$ at the point ( $a$ , a) cuts off intercepts $α$ and $β$ on the coordinate axes where $α^{2} + β^{2} = 61$ , then the value of $a$ is

1 25

2 36

3

\pm 30

4

\pm 40

Explanation:

(C) : Given, the curve -
$2 y^{3} = a x^{2} + x^{3}$
On differentiating both sides w.r.t. $x$ we get -
$6 y^{2} \frac{d y}{d x} = 2 a x + 3 x^{2}$
$\frac{d y}{d x} = \frac{2 a x + 3 x^{2}}{6 y^{2}}$
${(\frac{d y}{d x})}_{(a, a)} = \frac{5 a^{2}}{6 a^{2}}$
${(\frac{d y}{d x})}_{(a, a)} = \frac{5}{6}$
The equation of the tangent at $(a, a)$ is,
$y - a = \frac{5}{6} (x - a)$
$6 y - 6 a = 5 x - 5 a$
$5 x - 6 y + a = 0$
This intercepts lengths $= \frac{- a}{5}$ and $\frac{a}{6}$ with $x$ and $y$ axis respectively.
$∴ α = \frac{- a}{5}, β = \frac{a}{6}$
Now, $α^{2} + β^{2} = 61$
${(\frac{- a}{5})}^{2} + {(\frac{a}{6})}^{2} = 61$
$\frac{a^{2}}{25} + \frac{a^{2}}{36} = 61$
$\frac{36 a^{2} + 25 a^{2}}{25 \times 36} = 61$
$61 a^{2} = 61 \times 36 \times 25$
$a^{2} = 36 \times 25$
$a = 6 \times 5$
$a = \pm 30$
The value of a is $\pm 30$ .

COMEDK-2012

Application of Derivatives

85404 If the radius of a sphere is measured as $9 cm$ with an error of $0.03 cm$ , then find the approximating error in calculating its volume.

1

2.46 π {cm}^{3}

2

8.62 π {cm}^{3}

3

9.72 π {cm}^{3}

4

7.46 π {cm}^{3}

Explanation:

(C) : We have,
Radius of sphere $(r) = 9 cm$
and Error $Δ r = 0.03 cm$
Volume of sphere,
$V = \frac{4}{3} π r^{3}$
$\frac{dV}{dr} = 4 π r^{2}$
${(\frac{dV}{dr})}_{(r = 9)} = 4 π \times 9^{2} = 324 π$
Let, $Δ V$ be the error in $V$ due to error $Δ r$ in $r$ .
Then, $Δ V = \frac{dV}{dr} Δ r$
$Δ V = 324 π \times 0.03 = 9.72 π {cm}^{3}$
Thus, the approximate error in calculating the volume is $9.72 π {cm}^{3}$ .

COMEDK-2016